Lesson DB4-DB5
DB4-DB5 Travelling salesperson bounds and refining network models Quiz: AQA Further Maths, Unit 5
20 questions
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Lesson DB4-DB5, Travelling salesperson bounds and refining network models: 20 multiple choice questions for the AQA Further Maths (7367), Unit 5: Optional application 3: discrete mathematics, written with Revision Ninja.
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The 20 questions
-
What type of cycle does the travelling salesperson problem aim to find with minimum total weight?
- Maximum spanning tree
- Minimum Eulerian cycle
- Minimum Hamiltonian cycle
- Shortest path tree
-
Which algorithm produces a lower bound for the travelling salesperson problem?
- Deleted node MST
- Chinese postman algorithm
- Dijkstra's algorithm
- Nearest neighbour algorithm
-
Which algorithm gives an upper bound for the travelling salesperson problem?
- Nearest neighbour algorithm
- Kruskal's algorithm
- Prim's algorithm
- Deleted node MST
-
A TSP network has lower bound 18 and upper bound 20. What is the range for the optimal weight?
- At least 20
- At most 18
- 18 to 20
- Greater than 20
-
A complete network has nodes A, B, C, D, E with arc weights AB=4, AC=6, AD=5, AE=7, BC=3, BD=8, BE=6, CD=4, CE=5 and DE=2. Deleting node A leaves nodes B, C, D and E, whose minimum spanning tree has weight 9. What is the lower bound for the travelling salesperson problem?
- 20
- 9
- 18
- 13
-
A complete network has nodes A, B, C, D, E with arc weights AB=4, AC=6, AD=5, AE=7, BC=3, BD=8, BE=6, CD=4, CE=5 and DE=2. Using the nearest neighbour method starting at A, the tour is A-B-C-D-E-A. What is its total weight?
- 18
- 19
- 20
- 22
-
A complete network has nodes A, B, C, D, E with arc weights AB=4, AC=6, AD=5, AE=7, BC=3, BD=8, BE=6, CD=4, CE=5 and DE=2. What is the total weight of the tour A-B-C-E-D-A?
- 21
- 19
- 20
- 18
-
Removing one node and its two connected tour arcs leaves which structure on the remaining nodes?
- Complete graph
- Bipartite graph
- Spanning tree
- Hamiltonian cycle
-
How many distinct Hamiltonian cycles does a complete network on 6 nodes have, treating a cycle and its reverse as the same?
- 720
- 60
- 30
- 120
-
How can the upper bound from the nearest neighbour algorithm be improved for a TSP?
- Delete multiple nodes
- Reverse arc directions
- Remove longest edges
- Try different starts
-
A new arc is added to a TSP network model. How does the optimal tour weight change?
- Always decreases
- Decreases or unchanged
- Increases or unchanged
- Always increases
-
How is a one-way street represented when refining a network model?
- Loop arc
- Undirected edge
- Dummy node
- Directed arc
-
If deleting a node gives a minimum spanning tree of weight W on the remaining nodes, and the two smallest arcs at the deleted node have weights a and b, what is the lower bound for the travelling salesperson problem?
- W + a + b
- 2W + a + b
- W + a - b
- W + 2a + 2b
-
A TSP lower bound is 31 and upper bound is 38. What is the gap between these bounds?
- 12
- 35
- 69
- 7
-
To account for varying traffic levels during the day, how should arc weights be refined?
- Time-dependent weights
- Zero weights
- Negative weights
- Constant weights
-
Why is a TSP tour impossible in a network containing a node of degree 1?
- Requires degree zero
- Requires odd degree
- Requires degree two
- Requires degree three
-
Why can the deleted node MST lower bound be strictly less than the optimal TSP tour weight?
- Contains negative weights
- Not a tour
- Exceeds max capacity
- Has disconnected nodes
-
A complete network has nodes A, B, C, D, E with arc weights AB=4, AC=6, AD=5, AE=7, BC=3, BD=8, BE=6, CD=4, CE=5 and DE=2. Using the nearest neighbour method starting at C, the tour found is C-B-A-D-E-C. What is its total weight?
- 18
- 22
- 19
- 20
-
What primary benefit do lower and upper bounds provide for a TSP problem?
- Calculate exact weight
- Eliminate all nodes
- Guarantee unique route
- Bracket optimal weight
-
How many distinct Hamiltonian cycles does a complete network on 5 nodes have, treating a cycle and its reverse as the same?
- 120
- 24
- 6
- 12
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